Asked by valerie
A 7g bullet going at 300m/s hits and embeds itself into a 3.456kg block held on a frictionless table by a spring with k = 40N/m. Find the speed of the block and bullet after collision.
Answers
Answered by
Damon
Momentum before collision
= .007 * 300
= 2.1 kg m/s
Momentum after collision
= (3.456 +.007) v
= 3.463 v
The spring has not done anything yet so
momentum after = momentum before
3.463 v = 2.1
= .007 * 300
= 2.1 kg m/s
Momentum after collision
= (3.456 +.007) v
= 3.463 v
The spring has not done anything yet so
momentum after = momentum before
3.463 v = 2.1
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